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Practical Session 10

Computer Architecture and Assembly Language. Practical Session 10. Error detection and correction. data transmission via communication channels  channel noise  errors may happen we need techniques that enable reliable delivery of digital data over unreliable communication channels.

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Practical Session 10

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  1. Computer Architecture and Assembly Language Practical Session 10

  2. Error detection and correction • data transmission via communication channels channel noise  errors may happen • we need techniques that enable reliable delivery of digital data over unreliable communication channels • general idea: add some redundancy (extra data) to a message • this extra data is used • to check consistency of the delivered message • to recover data determined to be corrupted

  3. Hamming Distance Hamming distance (d) between two code words of equal length is anamount of 1-bit changes required to reach from one word to the other (number of 1’s in the XOR between the words) d(000, 011) = 2 000 011 10101 11110 d(10101, 11110) = 3 Minimum Hamming distance is thesmallest Hamming distance between all possible pairs in a set of words. Example: Given the following coding scheme, let’s find the minimum Hamming distance. 00000 01011 10101 11110 d(00000,01011)=3 d(00000,10101)=3 d(00000,11110)=4 d(01011,10101)=4 d(01011,11110)=3 d(10101,11110)=3 d=3

  4. Minimum Hamming Distance d can detectd-1 errors Example: 00000 01011 10101 11110 3 errors can take us from one legal code word to another can detect at least 3-1=2 errors d=3 01011 00000 01000 01010 01011 Minimum Hamming Distance d can fix d-1 erasures 3 erasures can take us from one legal code word to another can detect at least 3-1=2 erasures 01011 00000 0_000 0_0_0 0_0__ Minimum Hamming Distance d can fix errors 2 errors can take us from two different legal code words to the same illegal word 00000 01000 01010 11110 01110 can fix at least floor((3-1)/2)=1 error

  5. Given fourdata words, can we use 5-bit code words for fixing1 error? Answer: yes. We need a hamming distance d=3 to fix 1 error. 00000111001011101011 d(00000,11100)=3 d(11100, 10111)=3 d(00000,10111)=4 d(11100, 01011)=4 d(00000,01011)=3 d(10111, 01011)=3 We can create code words with independent graphs of a single error code words.

  6. Parity Check • Given a data word, a code word is created by adding a parity bit to the end of the data word • d=2 (why?) • Can detect 1 error and fix 1 erasure 0001 1 send: 00_1 1 0011 1 receive: erased bit is a parity of other bits: parity (0011) = 1 1 error occurs

  7. if p0 is wrong, we getp2 p1 p0 =√√x= 001b = 1p0 should be at index 1 if p1 is wrong, we getp1 p1 p0 =√x√= 010b = 2p1 should be at index 2 if p2 is wrong, we getp2 p1 p0 =x√√= 100b = 4p2 should be at index 4 if x0 is wrong, we getp2 p1 p0 =√xx= 011b = 3x0 should be at index 3 … if x3 is wrong, we getp2 p1 p0 =xxx= 111b = 7x3 should be at index 7 Hamming Code (4,3) We assume that there may be at most one error. 7 6 5 4 3 2 1 indexes: code word: 4 data bits: x0 x1 x2 x3 3 parity bits: p0 p1 p2 We calculate: why we interleave bits of data and parity in this way ? Example: We got 0100111 Sender sent 0000111 • p0 = 1 = x3 ⊕x1 ⊕x0 = 0 ⊕ 0 ⊕ 1 = 1 √ • p1 = 1 = x3 ⊕x2 ⊕x0 = 0 ⊕ 1 ⊕ 1 = 0 x • p2 = 0 = x3 ⊕x2 ⊕x1 = 0 ⊕ 1 ⊕ 0 = 1 x p2 p1 p0 =xx√ = 110b = 6 The error bit is at index 6 (from right to left) in the code word  error bit isx2

  8. Cyclic Redundancy Check • CRC is an error-detecting code • Given a data, its non trivial (cryptographic) challenge to find another data with the same CRC code.  CRC can protect data (even from deliberate damage)

  9. Binary Pattern as Polynomial

  10. CRC Polynomials (common)

  11. Sending – Calculation Steps • Generatoris chosen • sequence of bits, of which the first and last are 1 • CRC check sequence is computed by long-division of message by generator • check sequence has 1 fewer bits than generator • Check sequence is appendedto the original message

  12. Sending – Calculation Steps Compute 8-bit CRC a message ‘W’ (0x57). • Select Generator: CRC-8-ATM • x8 + x2 + x + 1 = 100000111 • ‘W’ is 01010111= x6 + x4 + x2 + x + 1 • Extend ‘W’ with 8 bits: 0101011100000000 • Perform XOR of the word get in step (3) by generator • CRC code is the remainder • Append CRC code to ‘W’

  13. Generating CRC Codelong-division of message by generator 0101011100000000 100000111 Each time apply XOR on the extended message, when place a generator from the left-most ‘1’ bit of the extended message xor 100000111 001011011000000 xor 100000111 0011010110000 xor 100000111 01010101100 xor 100000111 0010100010 • Stop when CRC code has 1 fewer bits than generator CRC Code: 10100010

  14. Receiving – Calculation Steps • Append CRC code to the message: 0101011110100010 • Perform long division by the generator • If the reminder is not 0: an error occurred 0101011110100010 100000111 xor 100000111 There is no errors in received message. result: 00000000

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